Prove that the product of a sphere and a torus is parallelizable

Question

The product of sphere and torus, $S^2\times \mathbb T^2$, is parallelizable. How to prove this?

Answer 1

Let $\tau$ be the tangent bundle of $S^2$. Observe that $\tau\oplus 1\cong 3$ (indeed, if we add to $\tau$ the normal bundle (for the standard embedding in $\mathbb R^3$) -- which is trivial -- we get a trivial vector bundle; $n$ denotes trivial $n$-dimensional bundle) and tangent bundle of $\mathbb T^2$ is trivial. Now let $\pi_1\colon S^2\times\mathbb T^2\to S^2$ and $\pi_2\colon S^2\times\mathbb T^2\to\mathbb T^2$ be natural projections; then $$T_{S^2\times \mathbb T^2}=\pi_1^*T_{S^2}\oplus\pi_2^*T_{\mathbb T^2}=\pi_1^*\tau\oplus\pi_2^*2=\pi_1^*\tau\oplus 2=\pi_1^*(\tau\oplus1\oplus1)=\pi_1^*4=4.$$

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Can't resist adding a bonus: a short and elementary proof that a product of spheres is parallelizable if one of them is odd (E.B.Staples, 1966).

Answer 2

It is sufficient to prove that $S^2\times T^1$ is parallelizable. This will imply the parallelizability of $S^2\times T^2$ being the product of two parallelizable manifolds.

The general problem of parallelizability of products of spheres was considered in Maurizio Parton's thesis:

1. As mentioned in the thesis introduction: the parallelizability of $S^2\times T^1$ is a special case of a theorem of M. Kervaire.

2. In the thesis, a new parallelizability proof was given for the more general case of $S^n\times T^1$ by explicit construction (Proposition 2.1.2).

The main idea is as follows:

Let $x = (x_i)$ be the Eucledian coordinates of $R^{n+1}$ and the sphere $S^n$ be given by:

$ |x|^2 = \sum_i x_i^2 = 1$

$S^n\times T^1$ is diffeomorphic to the quotient manifold $(R^{n=1}-0)/\Gamma$, where the group $\Gamma$ is generated by the map $x \mapsto e^{2\pi} x$. Then the projection:

$ R^{n=1}-0 \rightarrow S^n\times T^1$

$x \mapsto (x/|x|, \log|x| \mod 2\pi)$

is $\Gamma$ equivariant, thus defines a parallelization.